HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

Makematics · Jul 20, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep that's it.

===

$\int_0^1 x^m(1-x)^n \ dx$

from terry lee?

Sy123 · Jul 20, 2013

Re: MX2 Integration Marathon

Makematics said:
from terry lee?

I didn't get it from there, its just the Beta function lol, The recurrence relation is quite simple for one that has multiple variables so I thought I'd just ask it straight away.

Makematics · Jul 21, 2013

Re: MX2 Integration Marathon

Sy123 said:
I didn't get it from there, its just the Beta function lol, The recurrence relation is quite simple for one that has multiple variables so I thought I'd just ask it straight away.

o rite so terry got his from there.

Sy123 · Jul 21, 2013

Re: MX2 Integration Marathon

$\int_{\frac{1}{2}}^{2} \frac{\ln (x)}{1+x^2} \ dx$

Sy123 · Jul 22, 2013

Re: MX2 Integration Marathon

$\int_0^{\pi /2 } \frac{\sin^2 x}{\sqrt{x^2 - \frac{\pi}{2}x + \frac{\pi^2}{16} + 1}} \ dx$

Sy123 · Jul 23, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep.

==================================

$\int_2^{1+ \phi} \frac{(x+1)^3 (x-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} \ dx$

$$Where$ \ \phi = \frac{1+ \sqrt{5}}{2}$

Dividing both top and bottom by x^2 then x, knowing that x is positive in the domain of integration.

$\frac{(x+1)^3 (x-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} = \frac{\left(1- \frac{1}{x^2} \right)(x^2+2x+1)}{\frac{1}{x}(x^2+1)^2 \sqrt{x^2+ 1 + \frac{1}{x^2}}} = \frac{(x+2+ \frac{1}{x}) (1-\frac{1}{x^2})}{(x+\frac{1}{x})^2 \sqrt{x^2+ \frac{1}{x^2} + 1}}$

Then substitute in: $u= x + \frac{1}{x}$

Limits become 3 and 1.

$\int_1^3 \frac{(u+2) du}{u^2 \sqrt{u^2-1}}$

$u=\sec \theta$

Easy evaluation to get:

$\cos^{-1} \left( \frac{1}{3} \right ) + \frac{4\sqrt{2}}{3}$

Sy123 said:
$\int_{\frac{1}{2}}^{2} \frac{\ln (x)}{1+x^2} \ dx$

$u=\frac{1}{x}$

$I=\int_{1/2}^2 \frac{\ln x}{1+x^2} \ dx = \int_2^{1/2} \frac{-\ln(u)}{1 + \frac{1}{u^2}} \times \frac{-1}{u^2} \ du = -\int_{1/2}^2 \frac{\ln u}{1+u^2}$

$\therefore 2I =0 \ \Rightarrow I=0$

=======

$\int \frac{x^2-1}{x^4 - 2x^2 \cos(2\theta) + 1} \ dx$

asianese · Aug 24, 2013

Re: MX2 Integration Marathon

$\int \dfrac{x^4}{x^3-1}\mathop{\mathrm{d}x}$

Sy123 · Aug 24, 2013

Re: MX2 Integration Marathon

A cool result:

$\lim_{t\to 1} \int_{\sqrt{t}}^t \frac{x^3}{x^2- 1} \ dx$

integral95 · Aug 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi /2 } \frac{\sin^2 x}{\sqrt{x^2 - \frac{\pi}{2}x + \frac{\pi^2}{16} + 1}} \ dx$

I'm new to this so...
so you use that f(x) = f(a-x)
then you add them together.
2I = sin^-1(x-pi/4) then of course divide by 2

obliviousninja · Aug 24, 2013

Re: MX2 Integration Marathon

I'm a bit of newbie, but how do I insert equations in a reply

Sy123 · Aug 24, 2013

Re: MX2 Integration Marathon

MethewYan said:
I'm new to this so...
so you use that f(x) = f(a-x)
then you add them together.
2I = sin^-1(x-pi/4) then of course divide by 2

Yep

obliviousninja said:
I'm a bit of newbie, but how do I insert equations in a reply

You need to insert the LaTeX code into the tags [.tex] and [./tex] (without the dots)

For example if you quote this post to see coding used:

$Hello$ and $\frac{\pi }{\sqrt{2}}$
Commands such as \frac and \sqrt are used, a small guide is stickied in these forums

obliviousninja · Aug 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep

You need to insert the LaTeX code into the tags [.tex] and [./tex] (without the dots)

For example if you quote this post to see coding used:

$Hello$ and $\frac{\pi }{\sqrt{2}}$
Commands such as \frac and \sqrt are used, a small guide is stickied in these forums

Seems like a lot of effort, im probs better off doing it on paper then uploading a photo.

chris_power_96 · Aug 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
A cool result:

$\lim_{t\to 1} \int_{\sqrt{t}}^t \frac{x^3}{x^2- 1} \ dx$

Does it approach negative infinity?

asianese · Aug 24, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
Does it approach negative infinity?

$\frac{1}{2}\ln 2$

seanieg89 · Aug 24, 2013

Re: MX2 Integration Marathon

asianese said:
$\frac{1}{4}\ln 2?$

$\frac{1}{2}\log(2).$

(logs being natural of course.)

asianese · Aug 24, 2013

Re: MX2 Integration Marathon

Oops yes I can't factorise

Sy123 · Aug 24, 2013

Re: MX2 Integration Marathon

My friend conjectured:

$\lim_{t \to 1} \int_{\sqrt{t}}^{t} \frac{x^{n+1}}{x^n - 1} \ dx = \frac{1}{n} \ln(2)$

For integer n, it works for n=2, 3 and 19 by computation

Any ideas on how to prove/disprove this?

RealiseNothing · Aug 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
My friend conjectured:

$\lim_{t \to 1} \int_{\sqrt{t}}^{t} \frac{x^{n+1}}{x^n - 1} \ dx = \frac{1}{n} \ln(2)$

For integer n, it works for n=2, 3 and 19 by computation

Any ideas on how to prove/disprove this?

I've got it down to showing that:

$\lim_{t \to 1} \int_{\sqrt{t}}^{t} \frac{x}{x^n-1} dx = \frac{1}{n}\ln(2)$

seanieg89 · Aug 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
My friend conjectured:

$\lim_{t \to 1} \int_{\sqrt{t}}^{t} \frac{x^{n+1}}{x^n - 1} \ dx = \frac{1}{n} \ln(2)$

For integer n, it works for n=2, 3 and 19 by computation

Any ideas on how to prove/disprove this?

Oh that sounds fun, will have a look in a bit.

seanieg89 · Aug 25, 2013

Re: MX2 Integration Marathon

$\lim_{t\rightarrow 1}\int_{\sqrt{t}}^t \frac{x}{x^n-1}\, dx=\lim_{t\rightarrow 1}\int_{1/2}^1 \frac{t^{2s}\log(t)}{t^{ns}-1}\, ds =\int_{1/2}^1 \lim_{t\rightarrow 1} \left(\frac{t^{2s}\log(t)}{t^{ns}-1}\right)\, ds\\ = \int_{1/2}^1 \frac{ds}{ns} =\frac{1}{n}\log(2).$

(After interchanging the limit and the integral, we use L'Hospitals to evaluate the inner limit.)

HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

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